Circle Fundamentals
Arcs, Sectors & Central Angles
A central angle is the angle at the center of a circle. The arc length and sector area are proportional to the central angle.
| Measurement | Formula | Fraction of Circle |
|---|---|---|
| Arc Length | $L = \dfrac{\theta}{360} \times 2\pi r$ | $\dfrac{\theta}{360}$ of circumference |
| Sector Area | $A = \dfrac{\theta}{360} \times \pi r^2$ | $\dfrac{\theta}{360}$ of total area |
Inscribed vs Central Angles
An inscribed angle (vertex ON the circle) = HALF of the central angle that subtends the same arc.
Key Fact: Any inscribed angle subtending a diameter is exactly 90°. If a triangle is inscribed in a circle with one side being the diameter, the opposite angle is 90°.
Chord Properties & Tangent Lines
A perpendicular from the center to a chord bisects the chord. Use this with the Pythagorean theorem to find chord length.
A tangent line to a circle is perpendicular to the radius at the point of tangency. This creates a right angle — use Pythagoras.
10 Circle Traps
1. Area uses $r^2$; circumference uses $r$
$C = 2\pi r$ (linear); $A = \pi r^2$ (quadratic). Doubling $r$ doubles $C$ but quadruples $A$.
2. Arc formula uses central angle, not inscribed
The $\frac{\theta}{360}$ formula uses the CENTRAL angle, not an inscribed angle.
3. Inscribed angle = ½ central angle
If central angle = 80°, the inscribed angle subtending the same arc = 40°.
4. Tangent and radius are perpendicular
A tangent creates a 90° angle with the radius at the point of tangency — this creates a right triangle.
5. Semicircle inscribed angle = 90°
Any angle inscribed in a semicircle (subtending the diameter) is exactly 90°.
6. $\pi$ approximation issues
On GMAT, leave answers in terms of $\pi$ unless told to approximate. $\pi \approx 3.14$ or $\frac{22}{7}$.
7. Chord vs radius confusion
Not every chord is a radius. The diameter is the longest chord.
8. Sector area vs arc length
Both use $\frac{\theta}{360}$ but one multiplies by $\pi r^2$ (area) and the other by $2\pi r$ (length).
9. Equal arcs ≠ equal chords... wait
Equal arcs in the same circle DO produce equal chords, but in different circles they don't.
10. Concentric circles: difference of areas
Area between two concentric circles = $\pi R^2 - \pi r^2 = \pi(R^2-r^2)$.
10 GMAT Practice Questions
A circle has a circumference of $20\pi$. What is its area?
(B) $100\pi$. $C = 2\pi r = 20\pi$ → $r = 10$. Area = $\pi r^2 = 100\pi$.
A sector has a central angle of 90° in a circle of radius 6. What is the area of the sector?
(B) $9\pi$. Sector area = $\frac{90}{360} \times \pi(6)^2 = \frac{1}{4} \times 36\pi = 9\pi$.
A chord is 8 units long and is 3 units from the center of a circle. What is the radius?
(B) 5. The perpendicular from center bisects the chord. Half-chord = 4. $r^2 = 3^2 + 4^2 = 9+16=25$ → $r=5$.
A central angle of $120°$ subtends an arc. What angle does an inscribed angle subtending the same arc measure?
(B) 60°. Inscribed angle = ½ × central angle = ½ × 120° = 60°.
Two concentric circles have radii of 5 and 9. What is the area of the region between them?
(C) $56\pi$. Area between = $\pi(9)^2 - \pi(5)^2 = 81\pi - 25\pi = 56\pi$.
A tangent line from external point P touches a circle of radius 5 at point T. If PT = 12, what is the distance from P to the center O?
(C) 13. Radius OT is perpendicular to tangent PT. $PO^2 = PT^2 + OT^2 = 144 + 25 = 169$ → $PO = 13$.
If the radius of a circle is doubled, by what factor does the area increase?
(C) 4. $A = \pi r^2$. If $r$ doubles to $2r$: new area = $\pi(2r)^2 = 4\pi r^2 = 4 \times$ original area.
Is the area of circle C greater than 25π?
(1) The circumference of C is greater than 10π.
(2) The diameter of C is 11.
(D) Each alone sufficient. For area > 25π: need r > 5. (1): C > 10π → 2πr > 10π → r > 5. Area > 25π. Sufficient. (2): d = 11 → r = 5.5 > 5. Area = π(5.5)² = 30.25π > 25π. Sufficient. Both alone → (D).
A circle has area $36\pi$. What is the arc length of a $60°$ sector?
(A) $2\pi$. Wait — Area = $36\pi$ → $r = 6$. Arc length = $\frac{60}{360} \times 2\pi(6) = \frac{1}{6} \times 12\pi = 2\pi$. (A) $2\pi$ is correct.
An equilateral triangle is inscribed in a circle of radius 6. What is the side length of the triangle?
(C) $6\sqrt{3}$. For an equilateral triangle inscribed in a circle of radius $R$: side = $R\sqrt{3} = 6\sqrt{3}$.
Lesson Summary — Key Takeaways
Area = πr², Circumference = 2πr
Doubling the radius quadruples the area. These two formulas are foundation.
Arc/sector = (θ/360) × full measure
Both arc length and sector area use this fraction of the whole circle.
Inscribed angle = ½ central angle
Diameter subtends 180° at center → inscribed angle = 90°. Classic GMAT pattern.
Tangent ⊥ radius at point of tangency
Creates a right triangle. Use Pythagorean theorem to find missing distances.